Do I cast the result of malloc?
Sometimes I notice comments like that:
or
on questions where OP uses casting. The comments itself contain a hyperlink to this question.
That is in any possible manner inappropriate and incorrect as well. There is no right and no wrong when it is truly a matter of one's own coding-style.
It's based upon two reasons:
This question is indeed opinion-based. Technically, the question should have been closed as opinion-based years ago. A "Do I" or "Don't I" or equivalent "Should I" or "Shouldn't I" question, you just can't answer focused without an attitude of one's own opinion. One of the reason to close a question is because it "might lead to opinion-based answers" as it is well shown here.
Many answers (including the most apparent and accepted answer of @unwind) are either completely or almost entirely opinion-based (f.e. a mysterious "clutter" that would be added to your code if you do casting or repeating yourself would be bad) and show a clear and focused tendency to omit the cast. They argue about the redundancy of the cast on one side but also and even worse argue to solve a bug caused by a bug/failure of programming itself - to not #include <stdlib.h> if one want to use malloc().
I want to bring a true view of some points discussed, with less of my personal opinion. A few points need to be noted especially:
Such a very susceptible question to fall into one's own opinion needs an answer with neutral pros and cons. Not only cons or pros.
A good overview of pros and cons is listed in this answer:
https://stackoverflow.com/a/33047365/12139179
(I personally consider this because of that reason the best answer, so far.)
One reason which is encountered at most to reason the omission of the cast is that the cast might hide a bug.
If someone uses an implicit declared malloc() that returns int (implicit functions are gone from the standard since C99) and sizeof(int) != sizeof(int*), as shown in this question
Why does this code segfault on 64-bit architecture but work fine on 32-bit?
the cast would hide a bug.
While this is true, it only shows half of the story as the omission of the cast would only be a forward-bringing solution to an even bigger bug - not including stdlib.h when using malloc().
This will never be a serious issue, If you,
Use a compiler compliant to C99 or above (which is recommended and should be mandatory), and
Aren't so absent to forgot to include stdlib.h, when you want to use malloc() in your code, which is a huge bug itself.
Some people argue about C++ compliance of C code, as the cast is obliged in C++.
First of all to say in general: Compiling C code with a C++ compiler is not a good practice.
C and C++ are in fact two completely different languages with different semantics.
But If you really want/need to make C code compliant to C++ and vice versa use compiler switches instead of any cast.
Since the cast is with tendency declared as redundant or even harmful, I want to take a focus on these questions, which give good reasons why casting can be useful or even necessary:
Fact is, that the cast is redundant per the C standard (already since ANSI-C (C89/C90)) if the assigned pointer point to an object of fundamental alignment requirement (which includes the most of all objects).
You don't need to do the cast as the pointer is automatically aligned in this case:
"The order and contiguity of storage allocated by successive calls to the aligned_alloc, calloc, malloc, and realloc functions is unspecified. The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated)."
Source: C18, §7.22.3/1
"A fundamental alignment is a valid alignment less than or equal to
_Alignof (max_align_t). Fundamental alignments shall be supported by the implementation for objects of all storage durations. The alignment requirements of the following types shall be fundamental alignments:— all atomic, qualified, or unqualified basic types;
— all atomic, qualified, or unqualified enumerated types;
— all atomic, qualified, or unqualified pointer types;
— all array types whose element type has a fundamental alignment requirement;57)
— all types specified in Clause 7 as complete object types;
— all structure or union types all of whose elements have types with fundamental alignment requirements and none of whose elements have an alignment specifier specifying an alignment that is not a fundamental alignment.
- As specified in 6.2.1, the later declaration might hide the prior declaration."
Source: C18, §6.2.8/2
However, if you allocate memory for an implementation-defined object of extended alignment requirement, the cast would be needed.
An extended alignment is represented by an alignment greater than
_Alignof (max_align_t). It is implementation-defined whether any extended alignments are supported and the storage durations for which they are supported. A type having an extended alignment requirement is an over-aligned type.58)Source. C18, §6.2.8/3
Everything else is a matter of the specific use case and one's own opinion.
Please be careful how you educate yourself.
I recommend you to read all of the answers made so far carefully first (as well as their comments which may point at a failure) and then build your own opinion if you or if you not cast the result of malloc() at a specific case.
Please note:
There is no right and wrong answer to that question. It is a matter of style and you yourself decide which way you choose (if you aren't forced to by education or job of course). Please be aware of that and don't let trick you.
Last note: I voted to lately close this question as opinion-based, which is indeed needed since years. If you got the close/reopen privilege I would like to invite you to do so, too.
In C, you don't need to cast the return value of malloc. The pointer to void returned by malloc is automagically converted to the correct type. However, if you want your code to compile with a C++ compiler, a cast is needed. A preferred alternative among the community is to use the following:
int *sieve = malloc(sizeof *sieve * length);
which additionally frees you from having to worry about changing the right-hand side of the expression if ever you change the type of sieve.
Casts are bad, as people have pointed out. Especially pointer casts.
The concept behind void pointer is that it can be casted to any data type that is why malloc returns void. Also you must be aware of automatic typecasting. So it is not mandatory to cast the pointer though you must do it. It helps in keeping the code clean and helps debugging
For me, the take home and conclusion here is that casting malloc in C is totally NOT necessary but if you however cast, it wont affect malloc as malloc will still allocate to you your requested blessed memory space.
Another take home is the reason or one of the reasons people do casting and this is to enable them compile same program either in C or C++.
There may be other reasons but other reasons, almost certainly, would land you in serious trouble sooner or later.
As other stated, it is not needed for C, but for C++.
Including the cast may allow a C program or function to compile as C++.
In C it is unnecessary, as void * is automatically and safely promoted to any other pointer type.
But if you cast then, it can hide an error if you forgot to include stdlib.h. This can cause crashes (or, worse, not cause a crash until way later in some totally different part of the code).
Because stdlib.h contains the prototype for malloc is found. In the absence of a prototype for malloc, the standard requires that the C compiler assumes malloc returns an int. If there is no cast, a warning is issued when this integer is assigned to the pointer; however, with the cast, this warning is not produced, hiding a bug.
As others stated, it is not needed for C, but necessary for C++. If you think you are going to compile your C code with a C++ compiler, for whatever reasons, you can use a macro instead, like:
#ifdef __cplusplus
# define MALLOC(type) ((type *)malloc(sizeof(type)))
# define CALLOC(count, type) ((type *)calloc(count, sizeof(type)))
#else
# define MALLOC(type) (malloc(sizeof(type)))
# define CALLOC(count, type) (calloc(count, sizeof(type)))
#endif
# define FREE(pointer) free(pointer)
That way you can still write it in a very compact way:
int *sieve = MALLOC(int); // allocate single int => compare to stack int sieve = ???;
int *sieve_arr = CALLOC(4, int); // allocate 4 times size of int => compare to stack (int sieve_arr[4] = {0, 0, 0, 0};
// do something with the ptr or the value
FREE(sieve);
FREE(sieve_arr);
and it will compile for C and C++.
You don't cast the result of malloc, because doing so adds pointless clutter to your code.
The most common reason why people cast the result of malloc is because they are unsure about how the C language works. That's a warning sign: if you don't know how a particular language mechanism works, then don't take a guess. Look it up or ask on Stack Overflow.
Some comments:
A void pointer can be converted to/from any other pointer type without an explicit cast (C11 6.3.2.3 and 6.5.16.1).
C++ will however not allow an implicit cast between void* and another pointer type. So in C++, the cast would have been correct. But if you program in C++, you should use new and not malloc(). And you should never compile C code using a C++ compiler.
If you need to support both C and C++ with the same source code, use compiler switches to mark the differences. Do not attempt to sate both language standards with the same code, because they are not compatible.
If a C compiler cannot find a function because you forgot to include the header, you will get a compiler/linker error about that. So if you forgot to include <stdlib.h> that's no biggie, you won't be able to build your program.
On ancient compilers that follow a version of the standard which is more than 25 years old, forgetting to include <stdlib.h> would result in dangerous behavior. Because in that ancient standard, functions without a visible prototype implicitly converted the return type to int. Casting the result from malloc explicitly would then hide away this bug.
But that is really a non-issue. You aren't using a 25 years old computer, so why would you use a 25 years old compiler?
I put in the cast simply to show disapproval of the ugly hole in the type system, which allows code such as the following snippet to compile without diagnostics, even though no casts are used to bring about the bad conversion:
double d;
void *p = &d;
int *q = p;
I wish that didn't exist (and it doesn't in C++) and so I cast. It represents my taste, and my programming politics. I'm not only casting a pointer, but effectively, casting a ballot, and casting out demons of stupidity. If I can't actually cast out stupidity, then at least let me express the wish to do so with a gesture of protest.
In fact, a good practice is to wrap malloc (and friends) with functions that return unsigned char *, and basically never to use void * in your code. If you need a generic pointer-to-any-object, use a char * or unsigned char *, and have casts in both directions. The one relaxation that can be indulged, perhaps, is using functions like memset and memcpy without casts.
On the topic of casting and C++ compatibility, if you write your code so that it compiles as both C and C++ (in which case you have to cast the return value of malloc when assigning it to something other than void *), you can do a very helpful thing for yourself: you can use macros for casting which translate to C++ style casts when compiling as C++, but reduce to a C cast when compiling as C:
/* In a header somewhere */
#ifdef __cplusplus
#define strip_qual(TYPE, EXPR) (const_cast<TYPE>(EXPR))
#define convert(TYPE, EXPR) (static_cast<TYPE>(EXPR))
#define coerce(TYPE, EXPR) (reinterpret_cast<TYPE>(EXPR))
#else
#define strip_qual(TYPE, EXPR) ((TYPE) (EXPR))
#define convert(TYPE, EXPR) ((TYPE) (EXPR))
#define coerce(TYPE, EXPR) ((TYPE) (EXPR))
#endif
If you adhere to these macros, then a simple grep search of your code base for these identifiers will show you where all your casts are, so you can review whether any of them are incorrect.
Then, going forward, if you regularly compile the code with C++, it will enforce the use of an appropriate cast. For instance, if you use strip_qual just to remove a const or volatile, but the program changes in such a way that a type conversion is now involved, you will get a diagnostic, and you will have to use a combination of casts to get the desired conversion.
To help you adhere to these macros, the the GNU C++ (not C!) compiler has a beautiful feature: an optional diagnostic which is produced for all occurrences of C style casts.
-Wold-style-cast (C++ and Objective-C++ only)
Warn if an old-style (C-style) cast to a non-void type is used
within a C++ program. The new-style casts (dynamic_cast,
static_cast, reinterpret_cast, and const_cast) are less vulnerable
to unintended effects and much easier to search for.
If your C code compiles as C++, you can use this -Wold-style-cast option to find out all occurrences of the (type) casting syntax that may creep into the code, and follow up on these diagnostics by replacing it with an appropriate choice from among the above macros (or a combination, if necessary).
This treatment of conversions is the single largest standalone technical justification for working in a "Clean C": the combined C and C++ dialect, which in turn technically justifies casting the return value of malloc.
It depends on the programming language and compiler. If you use malloc in C, there is no need to type cast it, as it will automatically type cast. However, if you are using C++, then you should type cast because malloc will return a void* type.
This is what The GNU C Library Reference manual says:
You can store the result of
mallocinto any pointer variable without a cast, because ISO C automatically converts the typevoid *to another type of pointer when necessary. But the cast is necessary in contexts other than assignment operators or if you might want your code to run in traditional C.
And indeed the ISO C11 standard (p347) says so:
The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated)
malloc().In general, you don't cast to or from void *.
A typical reason given for not doing so is that failure to #include <stdlib.h> could go unnoticed. This isn't an issue anymore for a long time now as C99 made implicit function declarations illegal, so if your compiler conforms to at least C99, you will get a diagnostic message.
But there's a much stronger reason not to introduce unnecessary pointer casts:
In C, a pointer cast is almost always an error. This is because of the following rule (§6.5 p7 in N1570, the latest draft for C11):
An object shall have its stored value accessed only by an lvalue expression that has one of the following types:
— a type compatible with the effective type of the object,
— a qualified version of a type compatible with the effective type of the object,
— a type that is the signed or unsigned type corresponding to the effective type of the object,
— a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
— an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
— a character type.
This is also known as the strict aliasing rule. So the following code is undefined behavior:
long x = 5;
double *p = (double *)&x;
double y = *p;
And, sometimes surprisingly, the following is as well:
struct foo { int x; };
struct bar { int x; int y; };
struct bar b = { 1, 2};
struct foo *p = (struct foo *)&b;
int z = p->x;
Sometimes, you do need to cast pointers, but given the strict aliasing rule, you have to be very careful with it. So, any occurrence of a pointer cast in your code is a place you have to double-check for its validity. Therefore, you never write an unnecessary pointer cast.
In a nutshell: Because in C, any occurrence of a pointer cast should raise a red flag for code requiring special attention, you should never write unnecessary pointer casts.
Side notes:
There are cases where you actually need a cast to void *, e.g. if you want to print a pointer:
int x = 5;
printf("%p\n", (void *)&x);
The cast is necessary here, because printf() is a variadic function, so implicit conversions don't work.
In C++, the situation is different. Casting pointer types is somewhat common (and correct) when dealing with objects of derived classes. Therefore, it makes sense that in C++, the conversion to and from void * is not implicit. C++ has a whole set of different flavors of casting.
In C you can implicitly convert a void pointer to any other kind of pointer, so a cast is not necessary. Using one may suggest to the casual observer that there is some reason why one is needed, which may be misleading.
A void pointer is a generic pointer and C supports implicit conversion from a void pointer type to other types, so there is no need of explicitly typecasting it.
However, if you want the same code work perfectly compatible on a C++ platform, which does not support implicit conversion, you need to do the typecasting, so it all depends on usability.
Casting the value returned by malloc() is not necessary now, but I'd like to add one point that seems no one has pointed out:
In the ancient days, that is, before ANSI C provides the void * as the generic type of pointers, char * is the type for such usage. In that case, the cast can shut down the compiler warnings.
Reference: C FAQ
The casting of malloc is unnecessary in C but mandatory in C++.
Casting is unnecessary in C because of:
void * is automatically and safely promoted to any other pointer type in the case of C.<stdlib.h>. This can cause crashes.malloc is called and cast.On the other hand, casting may increase the portability of your program. i.e, it allows a C program or function to compile as C++.
I prefer to do the cast, but not manually. My favorite is using g_new and g_new0 macros from glib. If glib is not used, I would add similar macros. Those macros reduce code duplication without compromising type safety. If you get the type wrong, you would get an implicit cast between non-void pointers, which would cause a warning (error in C++). If you forget to include the header that defines g_new and g_new0, you would get an error. g_new and g_new0 both take the same arguments, unlike malloc that takes fewer arguments than calloc. Just add 0 to get zero-initialized memory. The code can be compiled with a C++ compiler without changes.
In C you get an implicit conversion from void * to any other (data) pointer.
From the Wikipedia:
Advantages to casting
Including the cast may allow a C program or function to compile as C++.
The cast allows for pre-1989 versions of malloc that originally returned a char *.
Casting can help the developer identify inconsistencies in type sizing should the destination pointer type change, particularly if the pointer is declared far from the malloc() call (although modern compilers and static analyzers can warn on such behaviour without requiring the cast).
Disadvantages to casting
Under the ANSI C standard, the cast is redundant.
Adding the cast may mask failure to include the header stdlib.h, in which the prototype for malloc is found. In the absence of a prototype for malloc, the standard requires that the C compiler assume malloc returns an int. If there is no cast, a warning is issued when this integer is assigned to the pointer; however, with the cast, this warning is not produced, hiding a bug. On certain architectures and data models (such as LP64 on 64-bit systems, where long and pointers are 64-bit and int is 32-bit), this error can actually result in undefined behaviour, as the implicitly declared malloc returns a 32-bit value whereas the actually defined function returns a 64-bit value. Depending on calling conventions and memory layout, this may result in stack smashing. This issue is less likely to go unnoticed in modern compilers, as they uniformly produce warnings that an undeclared function has been used, so a warning will still appear. For example, GCC's default behaviour is to show a warning that reads "incompatible implicit declaration of built-in function" regardless of whether the cast is present or not.
If the type of the pointer is changed at its declaration, one may also, need to change all lines where malloc is called and cast.
Although malloc without casting is preferred method and most experienced programmers choose it, you should use whichever you like having aware of the issues.
i.e: If you need to compile C program as C++ (Although it is a separate language) you must cast the result of use malloc.
The returned type is void*, which can be cast to the desired type of data pointer in order to be dereferenceable.
Casting is only for C++ not C.In case you are using a C++ compiler you better change it to C compiler.
In this question, someone suggested in a comment that I should not cast the result of malloc. i.e., I should do this:
int *sieve = malloc(sizeof(*sieve) * length);
rather than:
int *sieve = (int *) malloc(sizeof(*sieve) * length);
Why would this be the case?